Question #2e6dd

1 Answer
Mar 25, 2016

"3.30 g"

Explanation:

Start by comparing the final temperature of the water + aluminium sample with the initial temperatures of the two substances.

Notice that aluminium goes from an initial temperature of 100.0^@"C" to a final temperature of 26.1^@"C". Since the temperature of the metal decreases, you can say that the metal is losing heat.

On the other hand, the water goes from an initial temperature of 23.7^@"C" to a final temperature of 26.1^@"C". Since the temperature of the water is Increasing, you can say that it's gaining heat.

The idea here is that the heat lost by the metal will be equal to the heat gained by the water.

color(blue)(-q_"metal" = q_"water")" " " "color(orange)("(*)")

The Minus sign is used here because heat lost will always carry a negative sign.

Now, in order to determine how much heat was lost / gained, you can use the equation

color(blue)(|bar(ul(color(white)(a/a)q = m * c * DeltaTcolor(white)(a/a)|)))" ", where

q - the amount of heat gained / lost
m - the mass of the sample
c - the specific heat of the substance
DeltaT - the change in temperature, defined as the difference between the final temperature and the initial temperature

Water's specific heat can be found here

c_"water" = "4.18 J g"^(-1)""^@"C"^(-1)

https://en.wikipedia.org/wiki/Heat_capacity#Mass_heat_capacity_of_building_materials

So, you know the mass of water, its specific heat, and its change in temperature

DeltaT_"water" = 26.1^@"C" - 23.7^@"C" = 2.40^@"C"

so plug in your values and find how much heat was absorbed by the water

q_"water" = 21.9 color(red)(cancel(color(black)("g"))) * 4.18"J"/(color(red)(cancel(color(black)("g"))) * color(red)(cancel(color(black)(""^@"C")))) * 2.4color(red)(cancel(color(black)(""^@"C")))

q_"water" = "219.7 J"

This means that the aluminium must have lost the exact same amount of heat. Use equation color(orange)("(*)") to get

q_"metal" = -"219.7 J"

You know that you have

q_"metal" = m_"metal" * c_"metal" * DeltaT_"metal"

The change in temperature for the metal will be

DeltaT_"metal" = 26.1^@"C" - 100.0^@"C" = -73.9^@"C"

Rearrange the equation to solve for m_"metal" and plug in your values to find

m_"metal" = q_"metal"/(c_"metal" * DeltaT_"metal")

m_"metal" = (color(blue)(cancel(color(black)(-)))219.7color(red)(cancel(color(black)("J"))))/(0.900color(red)(cancel(color(black)("J")))/("g" * color(red)(cancel(color(black)(""^@"C")))) * (color(blue)(cancel(color(black)(-)))73.9color(red)(cancel(color(black)(""^@"C"))))) = color(green)(|bar(ul(color(white)(a/a)"3.30 g"color(white)(a/a)|)))

The answer is rounded to three sig figs.